Question

prove that cos 7 x +cos 5x ÷sin7x - sin5x =cotx​

Answer 1

Answer:

$\frac{cos \: 7x + cos \: 5x}{sin \: 7x - sin \: 5x} = cotx$

Step-by-step explanation:

$\frac{cos \: 7x + cos \: 5x}{sin \: 7x - sin \: 5x} = cotx \\ \\ on \: solving \: \: cos 7x+cos 5x \: \: \: separately \\ using \: cosx+cosy = 2cos \frac{x + y}{2} \: cos\frac{x - y}{2} \\ put \: x = 7x \: \: and \: \: y = 5x \\ = 2cos (\frac{7x + 5x}{2}) \: cos(\frac{7x - 5x}{2} )\\ = 2cos (\frac{12x}{2}) \: cos(\frac{2x}{2} ) \\ = 2 \: cos \:6x \: cosx \\ \\ on \: solving \: \: sin 7x - sin 5x \: \: \: separately \\ using \: sinx+siny = 2cos \frac{x + y}{2} \: sin\frac{x - y}{2} \\ put \: x = 7x \: \: and \: \: y = 5x \\ = 2cos (\frac{7x + 5x}{2}) \: sin(\frac{7x - 5x}{2} )\\ = 2cos (\frac{12x}{2}) \: sin(\frac{2x}{2} ) \\ = 2 \: cos \:6x \: sinx \\ \\ now \: \: \frac{cos \: 7x + cos \: 5x}{sin \: 7x - sin \: 5x} = \frac{2 \: cos \:6x \: cosx }{2 \: cos \:6x \: sinx } \\ = \frac{cosx}{sinx} \\ = cotx \\ hence \: proved$

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Answer 2

Answer:

cos7x + cos5x ÷ sin7x - sin5x

= 2 cos([7-5]/2)x × cos([7+5]/2)x ÷ 2cos([7+5]/2)x × sin([7-5]/2)x

{•.• cosa + cosb = 2cos([a-b]/2) × cos([a+b]/2)

sina + sinb = 2cos([a+b]/2) × sin([a-b]/2) }

=cos(2/2)x × cos(12/2)x / cos(12/2)x × sin(2/2)x

= cos(2/2)x / sin(2/2)x

= cotx